2015年新西兰数学奥林匹克第二题解答

新西兰数学奥林匹克总计3道题目, 考试时间为90分钟.

 

 

设 $$a,\ b,\ c,\ {a\over b} + {b \over c} + {c \over a},\ {a \over c} + {c \over b} + {b \over a}$$ 均为正整数. 证明: $a = b = c$.

解答:

易知若 $(a. b, c)\ne1$, 则可对 ${a\over b} + {b \over c} + {c \over a}$ 及 ${a \over c} + {c \over b} + {b \over a}$ 进行约分. 因此, 不失一般性, 假设 $(a, b, c) = 1$.

事实上, 若 $(a, b, c) = 1$, 则必有 $(a, b)= 1,\ (b, c) = 1,\ (c, a) = 1$. 理由是:

设 $p$ 是一个素数, 且有 $p\ \not|\ a$, $p^n\ \big{|}\big{|}\ (b, c)$ (即 $p^n\ \big{|}\ (b, c)$ 但 $p^{n+1}\ \not\big{|}\ (b, c)$).

令 $b = b_0p^n$, $c = c_0p^n$, ($p\ \not|\ (b_0, c_0)$). 则有 $${a\over b} + {b \over c} + {c \over a} = {a \over b_0p^n} + {b_0 \over c_0} + {c_0p^n \over a} = {a^2c_0 + ab_0^2p^n + b_0c_0^2p^{2n} \over ab_0c_0p^n}\in\mathbf{N^*}\Rightarrow p^n\ \big{|}\ c_0.$$ 同理可证, $p^n\ \big{|}\ b_0$. 因此 $p^n\ \big{|}\ (b_0, c_0)$ 与 $p\ \not|\ (b_0, c_0)$ 矛盾.

即 $(a, b, c) = 1\Rightarrow (a, b)= 1,\ (b, c) = 1,\ (c, a) = 1$ 得证.

若 $a, b, c$ 两两互素, 则有 $${a\over b} + {b \over c} + {c \over a} = {a^2c + b^2a + c^2b \over abc}\in\mathbf{N^*}\Rightarrow a\ \big{|}\ c^2b\Rightarrow a = 1.$$ 同理可得 $b= 1$, $c = 1$. 因此 $a = b =c$. 

Q.E.D.